Question: Compute $\sin(-60^\circ)$.
Rotating $60^\circ$ clockwise is the same as rotating $360^\circ - 60^\circ = 300^\circ$ counterclockwise, so $\sin(-60^\circ) = \sin (360^\circ - 60^\circ) = \sin 300^\circ$.

Let $P$ be the point on the unit circle that is $300^\circ$ counterclockwise from $(1,0)$, and let $D$ be the foot of the altitude from $P$ to the $x$-axis, as shown below.

[asy]

pair A,C,P,O,D;

draw((0,-1.2)--(0,1.2),p=black+1.2bp,Arrows(0.15cm));

draw((-1.2,0)--(1.2,0),p=black+1.2bp,Arrows(0.15cm));

A = (1,0);

O= (0,0);

label("$x$",(1.2,0),SE);

label("$y$",(0,1.2),NE);

P = rotate(300)*A;

D = foot(P,A,-A);

draw(O--P--D);

draw(rightanglemark(O,D,P,2));

draw(Circle(O,1));

label("$O$",O,NW);

label("$P$",P,SE);

//label("$A$",A,SE);

label("$D$",D,N);

[/asy]

Triangle $POD$ is a 30-60-90 triangle, so $DO = \frac{1}{2}$ and $DP = \frac{\sqrt{3}}{2}$.  Therefore, the coordinates of $P$ are $\left(\frac12,-\frac{\sqrt{3}}{2}\right)$, so $\sin(-60^\circ) = \sin300^\circ = \boxed{-\frac{\sqrt{3}}{2}}$.